How is the DQN loss derived from (or theoretically motivated by) the Bellman equation, and how is it related to the Q-learning update?
The Bellman equation in RL is usually defined $$v_\pi(s) = \sum_a \pi(a|s) \sum_{s', r} p(s', r|s, a)\left[r + v_\pi(s')\right] = \mathbb{E}_{s' \sim p, a \sim \pi}\left[r(s, a) + v_\pi(s')\right] \; .$$
The way you have written it is correct, but I just thought I would point this out. Regardless, your intuition is correct in that it expresses a recursive relationship such that the value of your current state $s$ is equal to the sum of the expected reward from this state plus the expected value of the state you transition into.
You do, in fact, implement the Q-learning update in Deep Q-Learning. The loss function that you minimise in DQN is
$$ L(\theta) = \mathbb{E}_{(s,a,r,s')\sim U(D)}\left[\left( r + \gamma \max_{a'}Q(s', a'; \theta^-) - Q(s, a; \theta)\right)^2 \right]\;$$
where $U(D)$ denotes uniformly at random from replay buffer $D$ and $\theta$ are your network parameters (the network parameterises the Q-function), and $\theta^-$ are a previous iteration of the parameters that are updated every $c$ episodes to help with convergence of the network.
As you can see, the loss function is minimising the 'Bellman error' error from your equation 2. Lets think about why this is.
The TD update you provide is gradually shifting the Q value for $(s, a)$ towards $r + \max_a Q(s', a)$ - this is what we want after all since it eventually converges to the optimal Q-function.
Now lets think about the Deep Q-learning case. We want our network to approximate $Q(s, a)$ and so if we train the network, using the MSE loss, with $r + \max_a Q(s', a)$ as our target then our network will gradually be shifted towards predicting $r + \max_aQ(s', a)$ (which again would give us optimal Q-values for state-action pairs), just like with the TD update.
This is assuming that you know how training of neural networks works so if you don't then I would recommend asking/searching for a relevant question that explains this.